gphilip

> I also get electrocuted easily when I use the escalator.

You get shocked easily when you use the escalator.You wouldn't be electrocuted more than once.

gphilip

4 anni fadiscuss

> Adding one edge after the exact number of edges required to create a Hamiltonian cycle (number of edges equal to number of vertices) would appear to break the property.

A Hamiltonian cycle is a simple cycle (no repeated vertices or edges) that contains every vertex of the graph. If a graph G has a Hamiltonian cycle, then adding more edges to G will not make that cycle go away; it will still be there. So the property of "has a Hamiltonian cycle" is not broken by adding more edges.

As a simple example: consider the graph which is the cycle on 5 vertices. That is, the graph has 5 vertices, and is just one big cycle with 5 edges. This graph has a Hamiltonian cycle (the entire graph itself is one such cycle). If we add an extra edge to this graph, say between vertices 1 and 3, the original Hamiltonian cycle does not go away.

> How can there be a lower bound other than zero? However small the possibility, surely given an infinite number of cases, there are infinite possibilities of a particular structure being created.

The lower bound can be other than zero because they are looking at the threshold (probability) at which the probability of the object existing goes from "very low" to "extremely high". This is alluded to in the following quote:

  "When edges are added to a random graph of N vertices with a probability of less than log(N)/N, for instance, the graph is unlikely to contain a Hamiltonian cycle. But when that probability is adjusted to be just a hair greater than log(N)/N, a Hamiltonian cycle becomes extremely likely."

I don't know the precise probabilities, but this would be something like: "When the probability of an edge being present is less than log(N)/N then the probability of there being a Hamiltonian cycle is 1/(N^2). When the probability of an edge being present is slightly more than log(N)/N then the probability of there being a Hamiltonian cycle becomes (1 - 1/(N^2))."

(Note that I plucked the above probabilities out of thin air just for the sake of illustration, just to give you an idea of the form that these statements take. For the precise probabilities, please ask Google.)

> What does this mean?

See: https://en.wikipedia.org/wiki/Sunflower_(mathematics)

> This seems to suggest multiple Hamiltonian cycles in a graph, contradicting the earlier definition that every vertex must be connected.

This is no contradiction. There can be multiple Hamiltonian cycles in a graph. Consider the complete graph on n vertices; there are roughly n-factorial-many Hamiltonian cycles. Any permutation of the vertices corresponds to one such cycle. Different permutations can correspond to the same cycle, so the number is not exactly n-factorial. But you get the idea.

gphilip

5 anni fadiscuss

I configured Fastmail to send emails that I compose from my work address, via the work SMTP server. Not any other type of email. And indeed, they don't send emails that I compose from my non-work addresses via my work SMTP server: they use their own server to send these.

The support person who is dealing with this also turned up the fact that this is a change that they made some time in 2019. Earlier to that, all emails from Fastmail Calendar were sent from the "[email protected]" address. They changed this in late 2019 so that Calendar now sends these mails from my work address (following some peculiar logic and email rewrite rules).

So yes, it is surprising that they send emails from my Calendar via my work SMTP server.

Submissions

Kinetic Bombardment

The Sahara swung between lush and desert conditions every 20k years (2019)

The Lisp Curse (2017)

Bring all your chats together

Easy Steps to Take Back Control of Your Privacy in 2022

Tell HN: Fastmail may be routing Calendar reminders via your work SMTP server

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